[EM] Re: completing Condorcet using ratings information

[Chris Benham]

James,

>>the ratings should be scaled and maximised among the members of  the
>>Schwartz set,
>>between steps  2  and 3.
>
>	Sorry, Chris, what exactly do you mean by "scaled and maximized"? Is that
>like raising the highest candidate to 100 and the lowest to 0?
>
CB: I should have written "maximised and scaled".  Yes,  the 
highest-rated  Schwartz-set member  is changed  to the maximum,
the lowest-rated member is changed to the minimum, and the middle-ranked 
are changed so as to keep their relative spots in the
range between those two extremes. For example if there are three 
Schwartz-set members then (assuming a rating scale of 0-100)
100, 99, 98  becomes  100, 50, 0; and  0, 1, 4  becomes  0, 25, 100; and 
 80, 20, 2 becomes 100, 23.684, 0.

26: Bush 100 > Dean 10 > Kerry 0
22: Bush 100 > Kerry 10 > Dean 0
26: Dean 100 > Kerry 90 > Bush 0
1: Dean 100 > Bush 50 > Kerry 0
21: Kerry 100 > Dean 90 > Bush 0
4: Kerry 100 > Bush 50 > Dean 0

In your example, the voters have thoughtfully done this themselves.
If this is not done, then I think there would remain some incentive to 
exaggerate ratings among the likelt top-tier contenders.

You wrote (Wed.Jun.9):

>Weighted pairwise comparison method
>
>Ballots:
>1. Ranked ballot. Equal rankings are allowed.
>2. Ratings ballot. e.g. 0-100, whole numbers only. Equal ratings allowed.
>Note: You can give two candidates equal ratings while still giving them
>unequal rankings. However, if you give one candidate a higher rating than
>another, then you must also give the higher-rated candidate a higher
>ranking.
> 
>Tally:
>1. Pairwise tally, using the ranked ballots only. Elect the Condorcet
>winner if one exists.
>	If no Condorcet winner exists:
>2. Determine the direction of the defeats using the ranked ballots for a
>pairwise comparison tally.
>3. Determine the strength of the defeats by finding the weighted magnitude
>as follows. We'll say that the particular defeat we're considering is
>candidate A beating candidate B. For each voter who ranks A over B, and
>*only* for voters who rank A over B, subtract their rating of B from their
>rating of A, to get the marginal utility. The sum of these winning
>marginal utilities is the total weighted magnitude of the defeat.
>4. Now that the directions of the pairwise defeats have been determined
>(in step 2) and the strength of the defeats have been determined (in step
>3), you can choose from a variety of Condorcet completion methods to
>determine the winner. Beatpath and ranked pairs are my preferred choices.
>
><end of definition>
>
CB: On reflection, better still would be to derive simple approval 
scores from the ballots by each ballot approving the
Schwartz-set member they rate above average (of  the members), and 
half-approving those they rate exactly average.
Then in step 3, the  "strength of the defeats" are the scores derived 
thus : on those ballots that rank the pairwise-winner
above the pairwise- loser, simple approvals for the winner minus simple 
approvals for the loser.
(Another good alternative  would be to simply eliminate the candidate 
with the lowest  approval score, and then  start again.)

One advantage of this version is that it should get rid of any need or 
demand for separate ranked-ballots. Voters would
have no disincentive to rate favoured candidates slightly differently, 
so the rankings can be simply inferred from the ratings.

In your example, all the candidates are in the Schwartz set. Converting 
the ballots to simple Approval ballots, we get:
48: Bush
26: Dean, Kerry
01: Dean>.5Bush
21: Kerry, Dean
04: Kerry>.5Bush
Approval scores-       Bush:50.5       Dean:48     Kerry:51

(From here I think all reasonable approaches elect Kerry. He has the 
highest approval score, and if we eliminate the candidate
with the lowest approval score (Dean), then he pairwise beats Bush.)
Proceeding with the more complicated version .  Bush pairwise beats 
Dean. On those ballots that rank Bush above Dean, Bush's
approval score is 50 and Deans is 0, so the "approval magnitude" of 
 Dean's defeat is 50.
Dean pairwise beats Kerry. On those ballots that rank Dean above Kerry, 
 Dean's approval score is 27 and Kerry's is 26 so the
"approval magnitude" of  Kerry's defeat is 1.
Kerry pairwise beats Bush. On those ballots that rank Kerry above Bush, 
Kerry's approval score is 51 and  Bush's approval
score is 2, so the "approval magnitude"  of   Bush's defeat is 49.
So now we have:
Bush>Dean  50
Kerry>Bush 49
Dean>Kerry  1

Kerry wins. MinMax drops the Kerry defeat. Ranked Pairs locks Bush>Dean 
and then Kerry>Bush and then skips Dean>Kerry
to give a final order  Kerry>Bush>Dean.

So that looks pretty good. What do you think?  

My old idea of completing Condorcet by compressing ranks also picks 
Kerry here, so I'm interested in  seeing how the two methods compare in 
future examples.

Chris Benham


>"Condorcet completed by Compressing Ranks:
>
>1.CR  ballots, voters give each candidate a score out of  100 ( or any 
>other round number that is much greater than the number of candidates.)
>
>2. Inferring the rankings from the ratings, elect the CW is there is 
>one.  If not, then eliminate the non-members of the Schwartz set.
>
>3.Those ballots which show as many preference-levels among the remaining 
>candidates as there are remaining candidates (N) are now converted to 
>ballots showing (N-1) preferences by disregarding the smallest gap in 
>the scores of adjacently ranked remaining candidates. If there is a tie 
>for "smallest gap" , then disregard the tied gap between the highest 
>adjacent preferences among the remaining candidates.
>
>4. Inferring rankings from these ballots (amended so that none show more 
>than N-1 preference-levels), elect the CW if there is one.  If not , 
>eliminate non-members of the Schwartz set, repeat step 3 and so on.
>
>
>An equivalent version which is easier to hand-count but less natural and 
>more awkward to vote is to have the voters rank the candidates and also 
>rank the the gaps in utility  between the adjacently-ranked candidates, 
>from largest to smallest."
>
>
>  
>


   


 



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